Lecture 2

Modulation and Multiplexing How to send data fast and far?

• 2-Values & Multi-Values Encoding, and Baud Rate & Bit Rate • Nyquist Theorem – • Shannon Theorem – Relationship between Speed & Bandwidth Relationship between Speed & Noise • Digital Encoding • Carrier, Modulation, Demodulation and Modem - Digital Modulations: FSK, ASK, PSK, QAM • Multiplexing and Demultiplexing - FDM (Frequency Division Multiplexing) - TDM (Time Division Multiplexing) - WDM (Wave Division Multiplexing) - CDMA (Code Division Multiple Access)

Lecture 2

Increase Digital Signal Transmission Speed

Pulse

(2-values) M=2, interval=T

0 1 0 0 1 0 0 …010010

Encoder Sender

T 2T 3T 4T 5T 6T

t

Pulse

(2-values) M=2, half T

0 0 1 0 0 1 0 T 3T

bit rate = 1/T unit: bps bits per second Transmission System/Channel

6T

t Increase bit rate by reducing T baud rate: pulses per sec.



Baud

= bps if M=2 Decoder Receiver Minimum T?

Pulse

(4-values) M=4, interval=T

0 1 0 0 1 0 1 1 1 0 0 T 2T 3T 4T 5T 6T

Increase bit rate by increasing M=2

M-values encoding 1 pulse = log 2

n M bits = n bits Maximum M?

1 Baud = n*bps

Lecture 2

Harry Nyquist

Basic Question: -- How many pulses could be transmitted per second, and recovered, through a channel/system of limited bandwidth B?

Nyquist’s Paper: -- Certain topics in telegraph transmission theory, Trans. AIEE, vol. 47, Apr. 1928 Born: February 7, 1889, Sweden Died: April 4, 1976,Texas, USA Institutions: Bell Laboratories, AT&T Known for -- Nyquist sampling theorem -- Nyquist rate -- Johnson–Nyquist noise -- Nyquist stability criterion -- Nyquist ISI criterion -- Nyquist filter Transmission System/Channel limited bandwidth

Lecture 2

Nyquist Theorem

Relationship between Transmission Speed and System Bandwidth

0 1 0 0 1 0

Data Transmission Speed Maximum Signal Rate: D

0

Encoder Sender

T 2T 3T 4T 5T 6T

t Transmission System/Channel Bandwidth=B Decoder Receiver

Nyquist Theorem:

1) Given a system/channel bandwidth B, the minimum T=1/2B, i.e., the maximum signal rate

D=2B pulses/sec (baud rate, Baud) = 2Blog 2 M bits/sec (bit rate, bps)

2) To transmit data in bit rate D, the minimum bandwidth of a system/channel must be

B>=D/2log 2 M (Hz)

Maximum M?

Explanations:

A hardware cannot change voltages so fast because of its physical limitation

Questions:

1) Assume a telephone channel bandwidth B=3000Hz and M=1024, what’s its maximum rate?

T F 2) Can we use the above channel to send a TV signal in real time? Why?

Lecture 2

Claude Shannon

Basic Question: -- How do bandwidth and noise affect the transmission rate at which information can be transmitted over an channel?

Shannon’s Paper: -- Communication in the presence of noise. Proc. Institute of RE. vol. 37, 1949 Born: April 30, 1916, Michigan Died: February 24, 2001, Massachusetts Fields: Mathematics & electronic engineering Institution: Bell Laboratories Known for -- Information theory -- Shannon–Fano coding -- Noisy channel coding theorem -- Computer chess, Cryptography . . . . . . Transmission System/Channel

Shannon Theorem

Relationship between Transmission Speed and Noise Lecture 2

0 1 0 0 1 0

t t Encoder Sender Transmission System/Channel Bandwidth=B s(t) + Decoder Receiver Maximum Signal Rate Data Transmission Speed

Channel Capacity

Noise n(t) S/N=s²(t)/n²(t) =10log 10 S/N (dB, decibel) called signal-to-noise ratio

Shannon Theorem:

1) Given a system/channel bandwidth B and signal-to-noise ratio S/N, the maximum value of M = (1+S/N) when baud rate equals B, and its channel capacity is,

C = Blog 2 (1+S/N) bits/sec (bps, bite rate)

2) To transmit data in bit rate D, the channel capacity of a system/channel must be

C>=D Two theorems give upper bounds of bit rates implement-able without giving implemental method.

Lecture 2

Channel Capacity

Nyquist-Shannon theorem C = Blog 2 (1+S/N) shows that the maximum rate or channel Capacity of a system/channel depends on bandwidth, signal energy and noise intensity. Thus, to increase the capacity, three possible ways are 1) increase bandwidth; 2) raise signal energy; 3) reduce noise

Examples

1.

For an extremely noise channel S/N  0, C  0, cannot send any data regardless of bandwidth 2.

3.

If S/N=1 (signal and noise in a same level), C=B The theoretical highest bit rate of a regular telephone line where B=3000Hz and S/N=35dB.

10log 10 (S/N)=35  log 2 (S/N)= 3.5x log 2 10 C= Blog 2 (1+S/N) =~ Blog 2 (S/N) =3000x3.5x log 2 10=34.86 Kbps If B is fixed, we have to increase signal-to-noise ration for increasing transmission rate.

Shannon theorem tell us that we cannot send data faster than the channel capacity, but we can send data through a channel at the rate near its capacity.

However, it has not told us any method to attain such transmission rate of the capacity.

Lecture 2

Digital Encoding

…010010110

Digital Encoder Sender Only short distance < 100m !

Encoding Schemes: - RZ (Return to Zero) - NRZ (Non-Return to Zero) # NRZ-I, NRZ-L (RS-232, RS-422) # AMI (ISDN) - Biphase # Manchester & D-Manchester (LAN) # B8ZS, HDB3 Digital Decoder Receiver Manchester encoding

Lecture 2

Carrier and Modulation

Important facts:

- The RS-232 connects two devices in a short distance (<15m). - It cannot be propagated far because its signal energy rapidly becomes weak with the increase of transmission distance.

- A sine wave can propagate farther. The sine wave is an analogy signal.

- A signal can be carried by the sine wave, called carrier, for long distance.

Carrier: Acos(2 - modify A

 πf c t +φ) where f c is called carrier frequency Modulation: change or modify values of A, f c , φ according to input signal s(t) A[s(t)]: Amplitude Modulation (AM)

- modify

f c

- modify φ

  f c [s(t)]: Frequency Modulation (FM) φ[s(t)]: Phase Modulation (PM) s(t) Modulator carrier Modulated Signal: m(t) = m[s(t), Acos(2 πf c t +φ)]

Acos(2

πf c t +φ)

Modulated Wave/Signal and Spectrum

Lecture 2 Original Signal Spectrum Carrier Frequency Single Signal Band

PM

Single Signal Band

Lecture 2 Digital Modulation input: digital signal output: analogy signal FSK – Frequency Shift Keying

Digital Modulation

Digital signal ASK modulated signal 2ASK ASK – Amplitude Shift Keying 2-ASK 0: A 1 cos2πf c t 1: A 2 cos2πf c t PSK – Phase Shift Keying 4-PSK 00: Acos(2πf c t+ 0 ) 01: Acos(2πf c t+ π/ ２ ) 10: Acos(2πf c t+ π ) 11: Acos(2πf c t+ 3π/2) PSK modulated signal 4PSK 0 0 1 0 1 0 1 1 DM Anim

QAM –

Quadrature Amplitude Modulation

Lecture 2

QAM: a combinational modulation of amplitude and phase

m(t) = A[s(t)] cos{2πf c t + φ[s(t)]} = p(t) cos(2πf c t) + q(t) sin(2πf c t) π/4 (90 ° ) phase difference between cos(x) and sin(x), called quadrature QAM is currently more common in digital communications 4-QAM, 8-QAM, 16-QAM, 32-QAM, 64-QAM, 128-QAM, 256-QAM, 512-QAM, …

101

.

sin 8-QAM

100

. .

.

..

011 110 111 010 000

. .

001

bit_rate = 3 x baud _rate cos 16-QAM

0101 1101 1100

.

.

.

0111 0100 1011

.

.

0011

sin

. .

. . .

0010 1111 0000 1010

.

.

.

.

.

.

1110 1001 1000 0110

bit_rate = 4 x baud _rate cos QAM Vid

Lecture 2

QAM Transmitter and Demo

m(t) m(t) = A[s(t)] cos{2πf c t + φ[s(t)]}

Lecture 2

Modulator, Demodulator and Modem

Modulator: accept bit sequence and modulate a carrier Demodulator: accepted a modulated signal, and recreated bit sequence Modem: a single device = modulator + demodulator

Lecture 2 Site 1 CompA1 CompB1 CompC1

How to send data efficiently?

3 Lines  Good?

Rate D a Rate D b Rate D c Site 2 CompA2 CompB2 CompC2 1 Line 1 Line

Lecture 2

Multiplexing, Multiplexer and Demultiplexer

Multiplexing is the set of techniques that allows simultaneous transmissions of multiple signals across a single data link.

3 lines  cost & inflexible CompA1 CompB1 CompC1 Rate D a Rate D b Rate D c CompA2 CompB2 CompC2 CompA1 CompB1 CompC1 D a D b D c Multiplexer 1 shared link: rate D D>=D a +D b +D c FDM, TDM, CDM D E M U X Demultiplexer CompA2 CompB2 CompC2

Lecture 2

FDM –

Frequency Division Multiplexing

FDM: - A set of signals are put in different frequency positions of a link/medium - Bandwidth of the link must be larger than a sum of signal bandwidths - Each signal is modulated using its own carrier frequency - Examples: radio, TV, telephone backbone, satellite, … f

A1 B1

1 2

C1

3

Mod

1

f

1

Mod

2

f

2

Mod

3

f

3 + 1

Dem

2 3

Dem Dem

1 2 3

A2 B2 C2

Lecture 2

TDM – Time Division Multiplexing

TDM: - Multiple data streams are sent in different time in single data link/medium - Data rate of the link must be larger than a sum of the multiple streams - Data streams take turn to transmit in a short interval - widely used in digital communication networks CompA1 CompB1 CompC1 … C1 B1 A1 C1 B1 A1 … D E M U X Anim1 , Anim2 CompA2 CompB2 CompC2

Examples of FDM and TDM

Lecture 2

FDM TDM

Lecture 2

Wave Division Multiplexing (WDM) and Spread Spectrum

WDM: - conceptually the same as FDM - using visible light signals (color division multiplexing) - sending multiple light waves across a single optical fiber Spread Spectrum: - spread the signal over a wider bandwidth for reliability and security - its carrier frequency is not fixed and dynamically changed - such changes is controlled by a pseudorandom 0/1 sequence (code) - the signal is represented in code-domain s(t) Code Mod Digital Mod

..0011001001010…

Pseudorandom code

Acos2

πf c t CDMA More CDMA (Code Division Multiple Access): different codes for different signals

8C32810.138ppt-Cimini-7/98

WIDEBAND CDMA (3G, NTT)

• The W-CDMA concept:

–

4.096 Mcps Direct Sequence CDMA

–

Variable spreading and multicode operation

–

Coherent in both up-and downlink = Codes with different spreading, giving 8-500 kbps ....

P f 4.4-5 MHz High rate multicode user Variable rate users 10 ms frame t

Exercise 2

1. Use Nyquist's Theorem to determine the maximum rate in bits per second at which data can be send across a transmission system that has a bandwidth of 4000 Hz and use four values of voltage to encode information. What's the maximum rate when encoding the information with 16 values of voltage? 2. Is it possible to increase a number of the encoded values without limit in order to increase transmission speed of system? Why? Assume a bandwidth of a system is 4000 Hz and a signal-to-noise ratio S/N=1023, What's the maximum rate of the system?

3. (True/false) A digital modulator using ASK, PSK or QAM is a digital-to-digital system. 4. (1) If the bit rate of 4-PSK signal is 2400bps, what’s its baud rate?

(2) If the baud rate of 256-QAM is 2400 baud, what’s its bit rate?

5. The bite rate of one digital telephone channel is 64Kbps. If a single mode optical fiber can transmit at 2 Gbps, how many telephone channel can be multiplexed to the fiber.

Assume TDM is used.